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Hailstones revisited

By

Joe Dickens and Sabrina Qian

Submitted by Marianne on July 9, 2010

"Mathematics is not yet ready for such problems." This is what the mathematician John Conway reportedly said about the Collatz conjecture, a simple-looking, yet unsolved mathematical mystery. This week Plus hosted two intrepid work experience students, Sabrina Qian and Joe Dickens, who decided to have a go at this fiendish problem nevertheless. And they made some intriguing discoveries. Here is what they came up with.

What are hailstone sequences and what's the Collatz conjecture?

Starting with any positive integer n, form a sequence in the following way:

If n is even, divide by 2 to get a new number n'= n/2

If n is odd, multiply by 3 and add 1 to get a new number n'=3n+1

.

Now take n' as your new starting number and repeat the process. For example,

A sequence formed in this way is known as a hailstone sequence because the values bounce up and down just like a hailstone in a cloud. You can use the applet below to generate hailstone sequences starting with any positive integer of your choice. Enter any positive integer, the hailstone sequence will be returned.

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It seems from our experiments that every hailstone sequence will eventually end up with the cycle 4, 2, 1, 4, 2, 1... and so on. Below we've plotted a few sequences starting with different numbers.

But can it be proved that every starting value will generate a sequence that ends up with the cycle 4, 2, 1? This is the unsolved problem known as the Collatz conjecture — no-one has so far been able to prove it.

What can we prove?

There are, however, some things we can prove. First of all, we can prove that 4, 2, 1 is the only simple recurring pattern. A simple recurring pattern is one that only involves one odd number, so that the transformation 3n+1 is only used once. In the example 4, 2, 1 the only odd number is 1 and the only time 3n+1 is needed is to transform 1 into 4.

Suppose a simple recurring pattern contains the odd number n. Then 3n+1 must be a multiple of 2n. This is because you must be able to return to the number n starting from 3n+1 using only division by 2. In other words, we have:

3n+1 = y2n,

where y is some positive integer.

Since y is a positive integer, the smallest it can be is 1. If y=1, then

3n+1= 2n.

This clearly cannot be true, because n is a positive integer, meaning that multiplying it by 3 will give a larger answer than multiplying it by 2. If we take the next lowest value of y, which is 2, we get the equation:

3n+1=4n.

Solving this for n gives

n=1.

This gives us the repeating pattern 1,4,2,1,4,2,... .

If we increase the value of y, so that y is at least 3, we get

3n+1 = y2n ≥ 6n,

which means that

1 ≥3n,

so

1/3 ≥ n.

This clearly cannot be true if n is a positive integer. Thus we have proved that 4, 2, 1 is the only simple recurring pattern. If a hailstone sequence eventually settles on a simple recurring pattern, then this pattern has to be 4, 2, 1,... .
However, this does not prove that there aren't recurring patterns involving more than one odd number.

Another thing that can be proved is that there are infinitely many numbers which, when "hailstoned", will produce the recurring pattern 4, 2, 1. In fact, any number n of the form

n=2m,

where m is a positive integer, will generate a hailstone sequence which eventually settles on the 4, 2, 1 cycle. Clearly such a number n is even, so the first step in the hailstoning process is a division by 2 to get 2m-1. If m-1=0, then 2m-1=1, so we have reached our 4,2,1 cycle. If m-1 > 0, then 2m-1 is even, so the next step in the hailstoning process is also a division by 2. Repeating this argument shows that all the numbers in the sequence are even, so all the steps in the process are divisions by 2, so we eventually get down to 4, then 2, then 1. This is true for any positive integer m, which proves that there are infinitely many numbers which when hailstoned eventually settle on our cycle. However, this doesn't prove that every number, when hailstoned, produces a sequence settling on 4, 2, 1. If this seems strange, here's a simple analogy: there are infinitely many even numbers, but that doesn't mean that every number is even.

Curiosities

When trying to start working on the problem, we looked into many areas of it, hoping to find some clues to help us when we tried to work out other things. One of the first areas we looked at was how many terms of hailstoning it took for the sequence to reach its first 4. At first the number of terms seemed fairly random, when starting with 5 it took 3 terms, when starting with 6 it took 6 terms, when starting with 7 it took 14 terms and when starting with 8 it took only one term. However, after some time it became apparent that there were some curious links between numbers.

The first thing that we noticed was that after a while some doubles began to appear. Doubles was the term we used when two consecutive numbers took the same number of terms to produce their first 4. The first example of this was when we hailstoned 12 and 13, which both took 7 terms to reach their first 4. This was followed by 14 and 15, which both took 15 terms to reach their first four. Although the distribution of the doubles seemed random, they consistently appeared, and before long there were also triples, quadruples and even quintuples! Below is a list of all of the doubles, triples, quadruples and quintuples that we found when hailstoning the numbers 1 to 184.

Another curiosity is that all of the doubles consist of an even number followed by an odd number (rather than the other way around), and that most of the doubles seem to come in small groups. It also seems strange that there are more quintuples than quadruples — surely it seems more likely to have 4 in a row than to have 5 in a row? And finally, why are there so many doubles, triples, quadruples and quintuples? If they appeared occasionally, they could be passed off as coincidence, but the fact that there are so many indicate that there is some underlying mathematical explanation. If you know what that might be, why not leave a comment on the Plusblog?

Comments

"But can it be proved that every starting value will generate a sequence that ends up with the cycle 4, 2, 1?" --

Hi,
Can the proof be something in this line?
We have already proved that for any n=2^m the sequence ends up with the cycle 4, 2, 1. Now need to prove that for any sequence S(n), where n is the seed of the sequence, there exists a subsequence S(m) belonging to S(n) such that m=2^p for any integer p. If n is odd then the next number in sequence is 3n+1=2q for any integer q. q can be even or odd. If q is odd then the next to next of 3n+1 in sequence is 3q+1 which is again even. If q is even it can be either written as 2^x for some integer x, in which case we have reached our subsequence or it can be 2y for some integer y, in which case we need to again proceed as above.
Following this line of logic I think we could prove our subsequence statement above. Does it work?

I noticed a similar pattern, however, I took it out to 10, 5, 16, 8, 4, 2, 1. Not every sequence ends with this pattern, however, as the numbers have to be large enough initially. It would be interesting to play around with though to determine if the pattern continues.
Sarah

Joe. The sequence 1,2,4,8,16,5,10,20 is not uniquely arrived at. 20 can only be reached from 40, but 40 can be reached from 80 or 13. Where there is a split possible, then both roads need to be taken into account. Then the conjecture is analogous.

Qualititatively it seems obvious that a hailstone sequence will eventually hit a value of 2^m. However, this is not a proof.

It would be interesting to see the "coverage" of numbers in hailstone sequences. eg

All even numbers will reduce eventually to an odd number. Hence we don't need to consider even numbers at all.

This leaves us with odd numbers. Odd numbers can be split into 2 types: 4n+1 and 4n-1.

The 4n+1 type has the following sequence

4n+1 --> 12n+4 --> 6n+2 --> 3n+1

This last number in the sequence is smaller than the first number - 4n+1 > 3n+1
So we can ignore 4n+1 numbers in our analysis because it will reduce to either another 4n+1 number (in which case it will reduce further) or it will reduce to a 4n-1 type.

Therefore the only types of odd number we need to consider are 4n-1 types.

eg 3, 7, 11, 15, etc.

I have some further analysis (eg 16m+7 always reduces when m is odd...), but I won't post just yet until I have more.

Hi
I've been thinking of something like this: if you can prove there is no other recurring pattern (of all kinds), it's obvious that any sequence ends up with 4,2,1 (eventually it will get to 4 and then it's stucked).

It's easy to prove there are no others positive recurring patterns length of 2,3,4 end even 5 numbers. I think that if you continue with this line you will maybe find the desirable proof :)

I realized that if the hailstone sequences of two consecutive numbers took the same amount of steps to reach 4, then they must start coinciding (by this I mean that the sequences become the same) at some point. For any two consecutive numbers that do this, lets call them n and n+1, their sequences coincide after m steps. Therefore every consecutive two numbers of the form n,n+1 (mod 2^m) are a double. Also, if you remove every consecutive two numbers of the form 4n+1, 4n+2, the pattern might become clearer, because if 4n+1, 4n+2 is a double, then 3n+1, 3n+2 are.

The hailstone sequencing is the same problem as the Josephus problem in reverse. We want there to be a sequence that doesn't follow the formula but the sequencing eventually hits a binary number or a number made of binary numbers which then make the solution ..4, 2, 1. If you subtract the highest binary number you can see how the base number will react... For example 10 eventually sheds its 8 and becomes a 2 1 which is again the binary sequencing we don't want. What is the issue mostly is that no matter how high the number it still eventually hits a binary number which is antiprime and logs down to 421... So let's take a number like 7;

7=>22 => 11 =>34. 34 is 32 + 2 which are both binary... A few more logs and we will reach 4;2;1. 17=> 52 =>26 =>13 => 40...20...10...=>1. As soon as you hit a decimal multiple of 10 with just the 4..2..1 you are about to finish the series. To see how many moves are left you can count the binary parts; subtract the smaller from the larger and then subtract that product from the whole and multiply the former product to the latter and you have the steps. It works for 10 but probably not for others; just a thought but hopefully I helped in some fashion.

There are problems that are easy to solve in theory, but impossible to solve in practice. Intrigued? Then join us on a journey through the world of complexity, all the way to the famous P versus NP conjecture.